The characteristic polynomial of a product

Let be an infinite field. Then, the affine space is irreducible in the Zariski topology (this follows since the ideal is prime): in other words, it is not the union of two proper algebraic subvarieties. This has the following consequence: if is a polynomial vanishing in a Zariski open set , it must vanish everywhere (since and so one of the terms in the right hand side must be ). Therefore, one may prove a polynomial identity by showing it holds on a Zariski open set.

For instance, let be matrices over . If , then obviously , since (in fact, we only need one of the two matrices to be invertible). Therefore, the identity , which is a polynomial in the entries of both and , holds in a Zariski open set (the set in given by pairs of matrices where one of them is invertible), and so by the previous observation holds for all square matrices.